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anyanavicka [17]
3 years ago
14

Differentiate between gravity andgravitation​

Physics
2 answers:
wariber [46]3 years ago
8 0

Answer:

Gravitation is the attractive force existing between any two objects that have mass. The force of gravitation pulls objects together. Gravity is the gravitational force that occurs between the earth and other bodies. Gravity is the force acting to pull objects toward the earth.

Explanation:

ZanzabumX [31]3 years ago
6 0
Gravitation is the attractive force existing between any two objects that have mass. The force of gravitation pulls objects together. Gravity is the gravitational force that occurs between the earth and other bodies. Gravity is the force acting to pull objects toward the earth.
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On your first day at work as an electrical technician, you are asked to determine the resistance per meter of a long piece of wi
Lostsunrise [7]

Answer:

0.06\Omega/m

Explanation:

Firstly, when you measure the voltage across the battery, you get the emf,

E = 13.0 V

In order to proceed we have to assume that the voltmeter offers no loading effect, which is a valid assumption since it has a very high resistance.

Secondly, the wires must be uniform. So the resistance per unit length is constant (say z). Now, even though the ammeter has very little resistance it cannot be ignored as it must be of comparable value/magnitude when compared to the wires. This is can seen in the two cases when currents were measured. Following Ohm's law and the resistance of a length of wire being proportional to it's length, we should have gotten half the current when measuring with the 40 m wire with respect to the 20 m wire (I=\frac{V}{R}). But this is not the case.

Let the resistance of the ammeter be r

Hence, using Ohm's law we get the following 2 equations:

\frac{13}{20z+r} =7.6   .......(1)

\frac{13}{40z+r} =4.5     ......(2)

Substituting the value of r from (2) in (1), we have,

13=152z+7.6\times\frac{13-180z}{4.5}

which simplifying gives us, z=0.0589\Omega/m\approx0.06\Omega/m (which is our required solution)

putting the value of z in either (1) or (2) gives us, r = 0.5325 \Omega

3 0
3 years ago
Michelle wants to Salt from water. She tried pouring the salt water through a piece of filter paper, but the salt did not separa
Katarina [22]
C. Let the water evaporate
3 0
3 years ago
HEY CAN ANYONE PLS PLS ANSWER DIS I RLY NEED IT
jeka94
Place the object in an electronic balance and measure its mass.
Place a measured amount of water in the cylinder.
Place the object in the cylinder so that it’s fully submerged.
Measure the new level of the liquid and subtract the original level. This is equal to the volume of the object.

Density = mass / volume.
8 0
3 years ago
Calculate the electric potential at point A, the middle of the rectangle, and at point B, the middle of the right-hand side of t
dsp73

Answer:

With the help of formula.

Explanation:

We can calculate the electric potential of any point through the formula of electric potential which is given below.

Electric potential =  Coulomb constant x charge/ distance of separation.

Symbolically it can be written as,  V = k q/ r where

V = electric potential  

k = Coulomb constant

q = charge

r = distance of separation

If we have all these data, we can simply put the data in the formula and we will get the value of electric potential.

5 0
3 years ago
Before beginning a long trip on a hot day, a driver inflates anautomobile tire to a gauge pressure of 2.70 atm at 300 K. At the
anygoal [31]

Answer:

The value is the temperature of the air inside the tire T_{2} = 340.54 K

% of the original mass of air in the tire should be released 99.706 %

Explanation:

Initial gauge pressure = 2.7 atm

Absolute pressure at inlet P_{1} = 2.7 + 1 = 3.7 atm

Absolute pressure at outlet P_{2} = 3.2 + 1 = 4.2 atm

Temperature at inlet T_{1} = 300 K

(a) Volume of the system is constant so  pressure is directly proportional to the temperature.

\frac{T_{2} }{T_{1} } = \frac{P_{2} }{P_{1} }

\frac{T_{2} }{300}  = \frac{4.2}{3.7}

T_{2} = 340.54 K

This is the value is the temperature of the air inside the tire

(b). Since volume of the tyre is constant & pressure reaches the original value.

From ideal gas equation P V = m R T

Since P , V & R is constant. So

m T = constant

m_{1}  T_{1} =  m_{2}  T_{2}

\frac{m_{2} }{m_{1}  } = \frac{T_{1} }{T_{2} }

\frac{m_{2} }{m_{1}  } = \frac{300}{354.54}

\frac{m_{2} }{m_{1}  } =0.00294

value of  the original mass of air in the tire should be released is  \frac{m_{2} - m_{1}}{m_{1}}

\frac{0.00294-1}{1}

⇒ -0.99706

% of the original mass of air in the tire should be released 99.706 %.

8 0
3 years ago
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